Supose that $X_1$ is a continuous and positive (real) random variable with exponential distribution, namely $$P(X_1>t)=e^{-\lambda t}\quad t>0$$ Now suppose that $X_2$ is another continuous and positive (real) random variable such that for every $s>0$ we have $$P(X_2>t\mid X_1=s)=P(X_1>t)=e^{-\lambda t}$$

Now, why can I conclude that $X_2$ has exponential distribution and moreover that $X_1$ and $X_2$ are independent?

The claim comes from the book "S. Ross - Stochastic Processes (second edition)" at page 64 when the author proves that the interrarival times in a Poisson process are i.i.d. random variables.

Thanks in advance


Since $e^{-\lambda t}$ does not depend on $s$, we have for all $t$ $$ P(X_2 > t) = \int_0^\infty \Pr(X_2 > t \mid X_1 = s) \lambda e^{-\lambda s}\,ds = e^{-\lambda t}. $$ Thus $X_2$ has exponential distribution with parameter $\lambda$ and the identity $$ \Pr(X_2 > t \mid X_1 = s) = \Pr(X_1 > t) = \Pr(X_2 > t), \qquad t,s > 0 $$ shows that $X_1$ and $X_2$ are independent.

  • $\begingroup$ Dubious: Are you actually able to prove that the identity implies the independence of $X_1$ and $X_2$? $\endgroup$ – Did Aug 26 '14 at 22:54
  • $\begingroup$ @Did: I guess I am missing something, but cannot we say that $\Pr(X_2 > t \mid X_1) = \Pr(X_2 > t)$, hence $\Pr(X_1 > t_1, X_2 > t_2) = E[1_{X_1 > t_1} \Pr(X_2 > t_2)] = \Pr(X_1 > t_1)\Pr(X_2 > t_2)$? $\endgroup$ – Siméon Aug 27 '14 at 8:01
  • $\begingroup$ Sure we can, but this argument relies on properties of conditional expectations in the full sense, which is why I asked @Dubious (not you) if they were able to complete the proof (with no reaction to this date, which I guess is a kind of answer to my question). $\endgroup$ – Did Aug 27 '14 at 8:06
  • $\begingroup$ @Did: ok! I thought that you were dubious about my statement... $\endgroup$ – Siméon Aug 27 '14 at 8:09
  • $\begingroup$ Of course not. Sorry that I was not more explicit about this. $\endgroup$ – Did Aug 27 '14 at 8:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.